计算机二级[英文]-C 字符串移动—东方仙盟

Use VC++2010 to open the solution in the candidate folder prog1. The project in this solution contains a source file prog1.c. Please write a function fun().

Function rule: Move the 1st ~ m-th characters to the end of the string. Move the characters from position m+1 to the end of the string to the front of the string.

For example: Original string: "ABCDEFGHIJK", $m=3$. After moving, the result string is "DEFGHIJKABC".

Notice: Do not modify the code inside main() and other functions. Only write code inside the braces of function fun().

复制代码

#include <stdio.h>
#include <string.h>
#define N 80
void fun(char *w, int m)
{

}
void main()
{
    FILE *wf;
    char a[N] = "ABCDEFGHIJK", b[N] = "ABCDEFGHIJK";
    int m;
    printf("The original string :\n");
    puts(a);
    printf("\n\nEnter m: ");
    scanf("%d",&m);
    fun(a,m);
    printf("\nThe string after moving :\n");
    puts(a);
    printf("\n\n");
/**********found**********/
    wf = fopen("out.dat","w");
    fun(b,3);
    fprintf(wf,"%s",b);
    fclose(wf);
/**********found**********/
}

Solution Idea

Get the total length len of string w.
We need to reorganize the string:
- First part: characters starting from index m to the end. (the latter part)
- Second part: characters from index 0 to m-1. (the first m chars)
We can use a temporary char array to store the new string, then copy it back to original string w.

Example: "ABCDEFGHIJK", m=3 len=11 part1: w $3$ ~ w $10$ → DEFGHIJK part2: w $0$ ~ w $2$ → ABC Combine → DEFGHIJKABC

Complete code for fun()

复制代码

void fun(char *w, int m)
{
    int len = strlen(w);
    char temp[N];
    int i, j = 0;
    // Copy m to end
    for(i = m; i < len; i++)
    {
        temp[j++] = w[i];
    }
    // Copy 0 to m-1
    for(i = 0; i < m; i++)
    {
        temp[j++] = w[i];
    }
    temp[j] = '\0'; // add string terminator
    strcpy(w, temp);
}

Another concise version (without temp array, using string library)

复制代码

void fun(char *w, int m)
{
    char t[N];
    strcpy(t, w+m);
    strcat(t, w);
    t[strlen(w)] = '\0';
    strcpy(w, t);
}

Explanation: w+m points to the m-th character. strcpy(t,w+m) gets the latter half; strcat appends the original first m characters, then truncate the string to original length.

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